Bc2641334137

International Journal of Modern Engineering Research (IJMER)

www.ijmer.com Vol. 2, Issue. 6, Nov-Dec. 2012 pp-4133-4137 ISSN: 2249-6645

www.ijmer.com4133 | Page

Serigiy L. Kryvyi1, Paweł Dymora

2, Mirosław Mazurek

3

1(Department of Distributed Systems, Rzeszow University of Technology, Poland)

ABSTRACT:The algorithm for computation of minimal

supported set of solutions and base solutions of linear

Diophantine equations systems in a ring of integer numbers

is proposed. This algorithm is founded on the modified TSS-

method.

Keywords:ring of integer numbers, linear Diophantine

equations, supported set, supported set of solutions

I. INTRODUCTION Linear Diophantine equations and also their

systems are often present in a wide variety of sciences with

heavy usage of computations. In order to solve many different systems these equations are brought to taskof

integerlinear programming, pattern

recognitionandmathematicalgames[2], cryptography[3],

unification[4], parallelizationof cycles[5], etc. In this case,

the sets of parameters of the equations are usuallysetsof

integers, residue ringor residue field of any number modulo

and sets in which solutions to the equations are found in

ringsof integers, the set ofnatural

numbersorfinitefieldsandresidue rings. Algorithms

forfinding solutionsoflinearDiophantineequations system

(SLDE) in the setof natural numbershave been described inmany publications [6] -[12]. In this work we will focus on

analyzes of the computation of SLDE algorithm in a ringof

integers. The basisof the proposedalgorithmis

theTSSmethodusedfor the constructing the minimalsetof

solutionsforming ahomogeneous system of

linearDiophantineequations(HSLDE) on thesetof natural

numbers[1].

II. PRELIMINARIES

The systems of linear Diophantine equations. The system

of linear Diophantine equations in a ring Z is described as

follows:

1 11 1 1 1

2 21 1 2 2

1 1

( ) ... ,

( ) ... ,

... ... ... ... ... ... ...

( ) ... ,

n n

n n

q q qn n q

L x a x a x b

L x a x a x bS

L x a x a x b

(1)

Where: , ,ij i ia b x Z , 1,...,i n , 1,...,j q . Solution to

SLDE (1) we will call a vector 1 2( , ,..., )nc c c c , which by

substitution in ( )iL x for the value jx , jc transforms

( )i iL c b into identity for all 1,2,...,i q . SLDE is called

Homogenous (HSLDE), where all ib are equal to 0,

otherwise the system is called inhomogeneous (ISLDE).

A. 2.1. The TSS method of HSLDE solution

Let’s consider HSLDE S presented as (1) and

1 (1,0,...,0)e , 2 (0,1,...,0)e , …, (0,...,0,1)ne which

are unitary vectors of canonical set base nZ . Let’s have M

as solutions set for the system S . Since it is homogeneous,

than the zero vectors is always valid solution. Such a

solution is called trivial and any other non- zero solution of

S is called non-trivial. The HSLDE is called contrary, only

when the set M is composed exclusively with the trivial

solution; otherwise it is called non contrary.

The TSS method and its implementation for linear

equations systems in a set of natural numbers have been

described in detail in [1]. Let’s consider a modification of

this method in case of the ring of integer numbers Z .

The case of homogeneous linear Diophantine equation

(HLDE). Let’s define the HLDE with the following form:

1 1( ) ... ... 0,i i n nL x a x a x a x (2) (2)

Where: , , 1,...,i ia x Z i n .

Let’s consider a set of canonical base vectors

1{ ,..., }nM e e and a function

1 1 2 2( ) ... n nL x a x a x a x HLDE (2). Without

limiting the generality, assume that in the function ( )L x the

first non-zero coefficient is 1a and 1 0a .

Let’s build a set of vectors

1 2 1 2 3 1{ ( , ,0,...,0), ( ,0, ,0,...,0),B e a a e a a

1 1 0( ,0,0,...,0, )}q qe a a M where 0 { : ( ) 0}r rM e L e ,

0ja , while, if for some ia GCD (Great Common

Divisor) 1 1( , ) 1a a , than the coordinates of this vector can

be reduced to this GCD. Selected non-zero coefficient 1a

will be called a primary. This way, one can assume, that all

vectors in the set B are such, that ia and 1a are mutually

simple. In other words, set B is constructed by combining

the first non-zero coefficient with the last non-zero

coefficient, having different signs and being complemented

with canonical base vectors, which correspond to zero

coefficients HLDE (2). This kind of constructed set is

called the TSS set or base set. It is obvious that vectors from the set B are solutions of HLDE (2), and the set B is

closed in a respect to summation, subtraction and

multiplication by an element from the ring Z .

Lemma 1. Let 1 2( , ,..., )qx c c c - be a some solution of

HLDE (2), than, if x B , than x can be represented as a

nonnegative linear combination in the form:

1 2 1 3 2 1... q qa x c e c e c e ,

where: , 1,..., 1ie B i q .

Proof.If 1( ,..., )qx c c M , than the vector has a following

representation:

The Computational Algorithm for Supported Solutions Set of

Linear Diophantine Equations Systems in a Ring of Integer

Numbers




1 2 1 3 2 1

2 1 3 3 2 1 1

1 1 2 1 1 1 1 2

...

( ... , ,..., )

( , ,..., ) ( , ,..., )

q q

q q q

q q

a x c e c e c e

c a c a c a c a c a

c a c a c a a c c c

Due to the fact that x is a solution of HLDE (2), i.e.

1 1 2 2 3 3 ... q qa c a c a c a c .

Note, that if a vector je from B is a canonical base vector

and the j -th coordinate of the vector x is equal to jc , than

in the vector x representation, the vector je enters with a

coefficient 1 ja c . Lemma proved.

The proved lemma results with the following conclusion.

Conclusion 1. If among the coefficients HLDE is even one coefficient equal 1, than the set B is a base of all HLDE

solutions set. Then indeed, the elements of the set B have

the form:

1 2 2

3 1

0

{ ( ,1,0, ...,0),

( ,0,1,0, ...,0),

( ,0,0, ...,0,1)}

q

q

e a e

a e

a M

,

i.e. if in the distribution of any solution x into vectors of

the set B the basic coefficient is equal one, then this

means, that the set B will be the base. Example 1. Let’s build TSS HLDE

1( ) 3 2 0L x x y z u v

The base set or the TSS base of the HLDE has the

following form:

1 2

3

4

( 1,3,0,0,0),

(1,0,3,0,0),

( 2,0,0,3,0),

( 1,0,0,0,3)

e e

e

e

The solutions LJRD 1 2(0,2,3,0,1), (1,1,0, 2,0)x x have

the representation 1 1 2 43 2 3x e e e , 2 1 33 2x e e .

All the solutions to any set of HLDE is presented as a set:

1 2

3

4

{ (1, 3,0,0,0),

(0,1,1,0,0),

(0, 2,0,1,0),

(0, 1,0,0,1)}

B e e

e

e

In this vector’s base 1x and 2x have the representation:

1 2 43x e e , 2 1 32x e e .

The case of homogenous system of linear Diophantine

equations. Let’s consider the HSLDE:

1 11 1 1

2 21 1 2

1 1

( ) ... 0,

( ) ... 0,

... ... ... ... ... ... ...

( ) ... 0,

n n

n n

q q qn n

L x a x a x

L x a x a xS

L x a x a x

Where: ,ij ia x Z , 1,...,i q , 1,...,j n .

Let’s build the base set 1 1 1

1 1 2 1{ , ,..., }qB e e e for the first

equation 1( ) 0L x and let’s calculate values 1

2 ( )i iL e b

where 1

1,i ie B b Z . Then, let’s create an equation:

1 1 1 1... ... 0i i q qb y b y b y (4) (4)

Then let’s build its base set '

1 1 2{ ,..., }qB s s . Vectors is

from '

iB corresponds to solutions vectors 2 2

2 1 2{ ,..., }qB e e

HSLDE 1 2( ) 0 ( ) 0L x L x .

Lemma 2. Vectors set B2 describes the base set of HSLDE

1 2( ) 0 ( ) 0L x L x , i.e. any solution x of this system has

a representation 2 2

1 1 2 2... q qkx l e l e , where: 2

2ie B ,

il Z , 1,..., 2i q .

Proof. Let’s have x to be any solution to HSLDE

1 2( ) 0 ( ) 0L x L x . Because x is a solution1( ) 0L x ,

and taking into account lemma 1, x can be represented as: 1 1

1 1 1 1... q qdx a e a e ,

Where: 1

1ie B , ia Z , 1,..., 1i q . Then, due to the fact

that x is the solution 2 ( ) 0L x we obtain:

2 1 1 1 1( ) ... 0q qL dx a b a b ,

Where1

2 ( )j jb L e , 1,..., 1j q . Respectively, vector

1 1( ,..., )qa a a is a solution of HLDE (4) and due to

lemma 1 we get:

1 1 2 2... q qka d s d s ,

Where: '

1is B , id Z , 1,..., 2i q , and k - is the main

coefficient of given HLDE . Thus: 2 2

1 1 2 2... q qkdx d e d e

where 2

2ie B , 1,..., 2i q .

The lemma 2 proved.

The following theorem can be proven with a help of mathematical induction, directly from lemma 1 and 2.

Theorem1. TSS HSLDE 1B (2) is built using the

described above manner and it is a base of all solutions set

of a given HSLDE.

Example 2. Let’s describe the base set of HSLDE:

1 1 2 3 4 5

2 1 2 3 4 5

( ) 3 2 0,

( ) 2 3 0 2 0.

L x x x x x xS

L x x x x x x

The base set for the first equation was described in a first

example. 1 1

1 1 2

1

3

1

4

{ (1, 3,0,0,0),

(0,1,1,0,0),

(0, 2,0,1,0),

(0, 1,0,0,1)}.

B e e

e

e

Values 2L x for a given vectors respectively equal to: -7,

3, -7, 1. let’s construct the equation

1 2 3 47 3 7 0y y y y and let’s transform the base set of

this HLDE: '

1 1 2 3{ (3,7,0,0), ( 1,0,1,0), ( 1,0,0,7)}.B s s s

These vectors correspond with TSS vectors (base set): 2 2

2 1 2

2

3

{ (3, 2,7,0,0),

( 1,1,0,1,0), ( 1, 4,0,0,7)}.

B e e

e

If during the construction of the base equation

1 2 3 47 3 7 0y y y y we perform combining in

accordance with the last value (ie. in respect to -1), we will

obtain such a base for a set of all solutions of a given

HSLDE:




2 2 2

1 2 3{ (1,4,0,0,7), (0, 2,1,0,3), (0,5,0,1, 7)}.e e e

In regards to conclusion 1, the theorem 1 result can

be detailed. Indeed, since the coefficients values in the

equation 1 11 1 12 2 1( ) ... 0n nL x a x a x a x are mutually

simple, it is always possible to have number one among the

values1( )L x . Without limiting the generality of

considerations we assume that GCD11 12 13( , , ) 1a a a , i.e.

the first three factors are mutually simple in 1( )L x . Then

there are such numbers1 2 3, ,d d d , that in vector

1 2 3( , , ,0,...,0)y d d d values 1( ) 1.L y once we obtain

this, let’s calculate the value 1( )L x for the canonical base

vectors. First let’s construct a base set 1B by combining the

spare vector y with the other vectors in order to obtain the

base set. Let’s note now that vectors from 1B have the

form:

1 11 1 2 12 2 3 13 3' , ' , ' ,e a y e e a y e e a y e

4 14 4 1' ,..., ' ,n n ne a y e e a y e

where ie – canonical base vectors;

ija – coefficient in an

equation 1( ) 0.L x

Vectors 'ie can be presented also in the following form:

1 11 1 11 2 11 3

2 12 1 12 2 12 3

3 13 1 13 2 13 3

4 14 1 14 2 14 3

' ( 1, , ,0,...,0),

' ( , 1, ,0,...,0),

' ( , , 1,0,...,0),

' ( , , ,1,...,0),

......................................................

e a d a d a d

e a d a d a d

e a d a d a d

e a d a d a d

1 1 1 2 1 3

.....

' ( , , ,0,...,1).n n n ne a d a d a d

In this form the following theorem take places.

Theorem 2. TSS HLDE 1B , is a set base of all solutions

for a given HLDE 1( ) 0L x and is constructed using the

method described above. The complexity of base

construction is proportional to the value 3l , where l - is a

maximal number of numbers m and n , n - the number of

unknowns in HLDE, and m - the maximum length of the

binary representation of HLDE coefficients.

Proof. Having 1 2( , ,..., )nx c c c - is the solution of HLDE

1( ) 0.L x Then vector x has the following representation:

1 1 2 2 3 3 4 4' ' ' ' ... 'n nx c e c e c e c e c e

1 11 1 1 2 12 1 3 13 1 4 14 1 1 1[( ... ],n nc a d c c a d c a d c a d c a d

1 11 2 2 12 2 2 3 13 2 4 14 2 1 2[ ... ],n nc c a d c a d c c a d c a d c a d

1 11 3 2 12 3 3 13 3 3 4 14 3

1 3 4

[

... ], ,..., )n n n

c a d c a d c a d c c a d

c a d c c

1 2 3 4( , , , ,..., )nc c c c c

because 11 1 12 2 1( ) ... 0.n nL x a c a c a c

The given algorithm complexity is described as a

complexity of the extended Euclidean algorithm, defined

along with the GCD and linear combination representing

this GCD. It is obvious (see [3]), that the complexity is

expressed as a value of ( log ),O m m where m - is the length

of the binary representation of the maximal HLDE

coefficient. Algorithm this is used no more than n times

and it is measured as a form of ( log ).O mn m Building the

base 1B requires no more than 3n operations. As a result,

the summary measure of the time complexity is described

as a value 3( ),O l where max( , ).l m n

The theorem is proved.

The above theorem leads to the following conclusion.

Conclusion 2. The time complexity of constructing all

solutions base set for the HSLDE with the form (5) is

proportional to a value 3( ),O ql where: q - is a number of

equations HSLDE, and max( , ).l m n

III. THE TSS METHOD OF ISLDE SOLUTION Having S be the ISLDE with the form of (1) and

0.qb Executing the free segments eliminations in the

first 1q equations, we transform the input ISLDE into the

following form:

1 11 1 1

2 21 1 2

1 11 1 1

1 1

' ( ) ' ... ' 0,

' ( ) ' ... ' 0,

...'

' ( ) ' ... ' 0,

' ( ) ' ... ' .

n n

n n

q q q n n

q q qn n q

L x a x a x

L x a x a x

S

L x a x a x

L x a x a x b

(5)

Let’s build the HSLDE base solutions set, composed of the

first 1q equations of the system (5). Having vectors

1{ ,..., }.ks s we specify ( ) , 1,..., .q j jL s a j k For this

values the following theorem is true.

Theorem 3. The ISLDE with the form (1) is consistent only

if the ISLDE 1 1 2 2 ... k k qa y a y a y b has at least one

solution in the set of integer numbers set.

Proof. If equation 1 1 2 2 ... k k qa y a y a y b has solution

1 2( , ,..., ),kc c c , then it is obvious that vector

1 1 2 2 ... k ks c s c s c s is a SNLRS solution.

If ISLDE is consistent and 1 2( , ,..., )ns k k k then its

solution s is presented in the linear combination form

constructed of the first 1q homogenous equations of the

system (5), i.e..:

1 1 2 2 ... .k ks c s c s c s

Then 1 1 2 2( ) ...q k k qL s c a c a c s b should has at least

one solution, because s is a solution of ISLDE.

The theorem is proved.

It is known that generalized solution of ISLDE has the form

1

,k

i i

i

y x a x

where: x – is a partial solution of ISLDE,

ix - is a base solution of a given HSLDE, ia - any integer

numbers, and k - is the number of base solutions. In this

case, for a comprehensive solution of the ISLDE we should

construct its HSLDE base and find one of the ISLDE

solutions. Finding such a solution, as a result from the

above considerations, is reduced into finding the solution of

the equation 1 1 2 2 ... .k k qa y a y a y b This solution can

be found with the use of the least coefficients method.

Example 3. The consistency of the ISLDE should be

checked:




1 1 2 3 4 5

2 1 2 3 4 5

( ) 2 3 0 1,

( ) 3 0 2.

L x x x x x xS

L x x x x x x

The transformed ISLDE has the following form:

1 1 2 3 4 5

2 1 2 3 4 5

' ( ) 7 5 3 2 0,'

( ) 3 0 2.

L x x x x x xS

L x x x x x x

The HLDE 1( ) ' 0L x base is combined with the vectors

(in this case the computation of GCD coefficients is not

necessary, because coefficient equals 1):

(1,0,0,0,7),(0,1,0,0, 5),(0,0,1,0,3),(0,0,0,1,2).

Values 2 ( )L x for these vectors equals: -4, 6, -2, -2. The

greatest common divisor of these values equals 2 and is a

divisor of the free member 2 2.b As a result the ISLDE

has a solution, this is it is consistent.

If the system is determined:

1 1 2 3 4 5

2 1 2 3 4 5

' ( ) 7 5 3 2 0,'

( ) 3 0 3,

L x x x x x xS

L x x x x x x

Therefore it has not solutions in a ring of integer numbers,

because GCD (-4, 6, -2, -2) = 2 and doesn’t divide the free

member -3 and then the equation 4 6 2 2 3x y z u

has no solutions.

In conclusion, we find that the given

measurements of the time complexity can be more detailed,

if we follow all the details of the processed calculations in

the TSS algorithm. In this work it is limited to determining

the polynominality of these algorithms.

IV. THE EXAMPLE USE OF THE ALGORITHM IN THE

INTERCONNECTION NETWORK DESIGNING

PROCESS Let’s consider the problem of designing the

interconnection network with the configuration presented in

the Fig. 1.

Fig. 1. The designed interconnection network configuration

schema.

The designed interconnection network has 10 inputs

1 10x x and two panels with outputs 1 3y y and 4 7y y

respectively. We should power supply the minimal amount

of inputs, in order to satisfy the following conditions:

1 10y , 2 20y , 3 5y , 4 20y , 5 50y , 6 10y ,

7 20y [V]. To solve this problem we construct the

following equations system:

1 2 3 9

3 4 5 10

2 6 7 9

1 3 7 10

3 5 6 9

1 4 5 7 8 9

2 3 4 6 7 8

10

20

5

20

20

10

20

x x x x

x x x x

x x x x

x x x x

x x x x

x x x x x x

x x x x x x

Such equations system has the following solutions: 0 (0,80, 50, 35,5, 45,0,0, 40,30)x . Therefore, we

obtain that: 1 0x ,

2 80x ,3 50x ,

4 35x ,5 5x ,

6 45x ,7 0x ,

8 0x ,9 40x ,

10 30x [V]. This means

that inputs 1 7 8, ,x x x may not be connected to any outputs.

The solutions of the homogenous equations system, which

corresponds to a given inhomogeneous equations system

may be described as follows:

( 1,11, 6,21, 3, 7, 23,46, 4,30)e ,

( 1,11, 6,22, 3, 7, 24,48, 4,31)s ,

(8, 90,49, 175,25,54,192, 383,33, 249)t ,

(5, 169,92, 329,47,107,361, 720,62, 468)r .

Using these solutions the interconnection network designer

has a choice, because he can select a special variant which

suits him best. Thus, the interconnection network designer

can use the general solution equation of a given equations system:

0 ,x x ae bs ct dr

Where , , ,a b c d are arbitrary integers?

V. CONCLUSION

In this paper we have presented the algorithm for

computation the minimal supported set of solutions and

base solutions of linear Diophantine equations systems in a

ring of integer numbers. Linear Diophantine equations and

their systems are often found in a wide variety of sciences

which have heavy usage of computations. Solving the Diophantine equations is one of the main issues in

computing the data, dependences in algorithm’s code

especially nested loop programs with memory access which

very often occurs in numerical computing. Recently, the

Diophantine equations are used to obtain an accurate and

predictable computational model in many multi-disciplinary

scientific fields especially in bimolecular networks studies.

The most crucial task in such models is to check

the model’s correctness – model validation problem. In the

validation process finding of basic state equations is the

most important task which may be checked by existence of

integer solutions of Diophantine equations systems. Moreover, one of the motivations of this problem comes

from the coding theory which may be implemented in many

different fields of cryptography, encryption and users

authentication (symmetric-key encryption and public-key

cryptosystems) where the nonuniqueness of Diophantine

equations solutions is analyzed. Thus, the proposed

research is an actual scientific problem.




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